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kvv77 [185]
2 years ago
7

It's in the picture ?

Mathematics
1 answer:
nignag [31]2 years ago
4 0

Answer:

<h2>4</h2>

Step-by-step explanation:

From first piece -6 < x ≤ 0.

From second piece 0 < x ≤ 4.

Therefore -6 < x ≤ 4 → x ∈ (-6, 4]

-7 ∉ (-6, 4]

-6 ∉ (-6, 4]

4 ∈ (-6, 4]

5 ∉ (-6, 4]

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Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y
ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; (x^2-4)^2y'' + (x + 2)y' + 7y = 0

The above equation can be better expressed as:

y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

p(x) = \dfrac{(x+2)}{(x^2-4)^2} \

p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \

p(x) = \dfrac{1}{(x+2)(x-2)^2}

Also;

q(x) = \dfrac{7}{(x^2-4)^2}

q(x) = \dfrac{7}{(x+2)^2(x-2)^2}

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

\lim \limits_{x \to-2} (x+ 2) p(x) =  \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{1}{(x-2)^2}

\implies \dfrac{1}{16}

\lim \limits_{x \to-2} (x+ 2)^2 q(x) =  \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}

\implies  \lim \limits_{x \to2}  \dfrac{7}{(x-2)^2}

\implies \dfrac{7}{16}

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0
3 years ago
Given that , the _____ justifies the conclusion that ∆ABD ≅ ∆AEC.
trasher [3.6K]
AAS Postulate

It is given that CE = BD so we know "S" (representing side) has to be in the three letter postulate.
It is also given that angle DBA and angle CEA are right angles, so therefore they are congruent. Now we know that an "A" must also be in the postulate.

Lastly, we know that the triangles have a second angle, EAB, in common because they share it overlappingly. So there must be another "A" in the postulate. 

Now we need to look at the order in which it is presented. The order follows Angle, Angle, Side so the postulate must be the AAS postulate. Hope this helps!

3 0
3 years ago
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Where's the drawing? 
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2 years ago
Solve for the value of n<br> (9n)<br> (8n+8)
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Answer:

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2 years ago
The length of a rectangle is 4 times the width. the area is 256 square centimeters. find the length and width of the rectangle.
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Width=8
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Look at the GCF of 256
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